a, khi x=2, ta đc:

Q=\(\frac{2+1}{2^2-9}\) ​
=\(\frac{3}{-5}\)

b, \(P = \frac{2 x^{2} - 1}{x^{2} + x} - \frac{x - 1}{x} + \frac{3}{x + 1}\)

 \(=\frac{2 x^{2} - 1}{x \left(\right. x + 1 \left.\right)}-\frac{\left(\right. x - 1 \left.\right) \left(\right. x + 1 \left.\right)}{x \left(\right. x + 1 \left.\right)}+\frac{3 x}{x \left(\right. x + 1 \left.\right)}\)

\(=\frac{2 x^{2} - 1 - \left(\right. x^{2} - 1 \left.\right) + 3 x}{x \left(\right. x + 1 \left.\right)}\)

\(=\frac{2 x^{2} - 1 - x^{2} + 1 + 3 x}{x \left(\right. x + 1 \left.\right)}\)

\(=\frac{x^{2} + 3 x}{x \left(\right. x + 1 \left.\right)}=\frac{x + 3}{x + 1}\).

c) Ta có

 \(M = P . Q = \frac{x + 3}{x + 1} . \frac{x + 1}{x^{2} - 9} = \frac{x + 3}{\left(\right. x - 3 \left.\right) \left(\right. x + 3 \left.\right)} = \frac{1}{x - 3}\)

Vì \(M=\frac{- 1}{2}\) suy ra \(\frac{1}{x - 3} = \frac{- 1}{2}\)

\(x - 3 = - 2\)

\(x = 1\).

a, khi x=2, ta đc:

Q=\(\frac{2+1}{2^2-9}\) ​
=\(\frac{3}{-5}\)

b, \(P = \frac{2 x^{2} - 1}{x^{2} + x} - \frac{x - 1}{x} + \frac{3}{x + 1}\)

 \(=\frac{2 x^{2} - 1}{x \left(\right. x + 1 \left.\right)}-\frac{\left(\right. x - 1 \left.\right) \left(\right. x + 1 \left.\right)}{x \left(\right. x + 1 \left.\right)}+\frac{3 x}{x \left(\right. x + 1 \left.\right)}\)

\(=\frac{2 x^{2} - 1 - \left(\right. x^{2} - 1 \left.\right) + 3 x}{x \left(\right. x + 1 \left.\right)}\)

\(=\frac{2 x^{2} - 1 - x^{2} + 1 + 3 x}{x \left(\right. x + 1 \left.\right)}\)

\(=\frac{x^{2} + 3 x}{x \left(\right. x + 1 \left.\right)}=\frac{x + 3}{x + 1}\).

c) Ta có

 \(M = P . Q = \frac{x + 3}{x + 1} . \frac{x + 1}{x^{2} - 9} = \frac{x + 3}{\left(\right. x - 3 \left.\right) \left(\right. x + 3 \left.\right)} = \frac{1}{x - 3}\)

Vì \(M=\frac{- 1}{2}\) suy ra \(\frac{1}{x - 3} = \frac{- 1}{2}\)

\(x - 3 = - 2\)

\(x = 1\).

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